So I was looking around for some programming information and found Project Euler. They have a number of problems to program. You can choose any programming language, it’s just to enhance your programming abilities.

So here is the answer to problem number one.

#!/usr/bin/perl

use warnings;
use strict;

my $sum;
my $count;

for ($count = 0; $count < 1000; $count ++) {
        if (($count % 3 == 0) || ($count % 5 == 0)) {
                $sum = $count + $sum;
                print "\$sum is $sum\n"; }
        }

It’s a really simple exercise. You just need to sum of all the multiples of 3 or 5 below 1000.

So I was thinking about writing a script to calculate the savings over a certain matter of time and came up with the following.

#!/usr/bin/perl

use warnings;
use strict;

print "What is the yield?\n";
my $yield = <>;
print "How many years?\n";
my $years = <>;
print "How much money saved anually?\n";
my $savings = <>;
my $i;
my $new_savings = 0;

for($i = 1; $i <= $years; $i++) {
	$new_savings = $savings + $new_savings;
	$new_savings = $new_savings + $new_savings*($yield/100);
	printf "Savings for year $i are \$%.2f.\n", $new_savings;
}

Below is an usage example.

[19:43:04] xavi@ubuntu:/tmp $ ./yield.pl
What is the yield?
5.0
How many years?
5
How much money saved anually?
10000
Savings for year 1 are $10500.00.
Savings for year 2 are $21525.00.
Savings for year 3 are $33101.25.
Savings for year 4 are $45256.31.
Savings for year 5 are $58019.13.
[19:43:18] xavi@ubuntu:/tmp $

This just gives an idea of the money that can be saved over a short period of time with a five percent yield. It’s a really simple script.